Exercise (8.4): Solutions - Angle Bisector Theorem

Angle Bisector Theorem
A	The bisector of an interior angle of a triangle divides the opposite side internally into a ratio equal to the ratio of the other two sides of the triangle. တြိဂံတစ်ခု၏ အတွင်းထောင့်တစ်ခုကို ထက်ဝက်ပိုင်းသော မျဉ်းသည် မျက်နှာချင်းဆိုင်အနားကို အတွင်းပိုင်းမှ ပိုင်းဖြတ်ရာ ပိုင်းဖြတ်လိုက်သော (အတွင်းပိုင်း) အချိုးသည် ကျန်အနားနှစ်ဖက် အချိုးနှင့် ညီသည်။
B	The bisector of an exterior angle of a triangle divides the opposite side externally into a ratio equal to the ratio of the other two sides of the triangle. တြိဂံတစ်ခု၏ အပြင်ထောင့်တစ်ခုကို ထက်ဝက်ပိုင်းသော မျဉ်းသည် မျက်နှာချင်းဆိုင် အနားကို အပြင်ပိုင်းမှ ပိုင်းဖြတ်ရာ ပိုင်းဖြတ်လိုက်သော (အပြင်ပိုင်း) အချိုးသည် ကျန်အနားနှစ်ဖက် အချိုးနှင့် ညီသည်။

1. Which of the following proportions follow from the fact that $AE$ bisects $\angle WAV$ in $\triangle WAV$ ?
   
   (a) $ \displaystyle \frac{W E}{E V}=\displaystyle \frac{W A}{A V}$
   
   (b) $ \displaystyle \frac{W E}{E V}=\displaystyle \frac{V A}{A W}$
   
   (c) $ \displaystyle \frac{W E}{W A}=\displaystyle \frac{E V}{A V}$
   
   (d) $ \displaystyle \frac{A V}{A W}=\displaystyle \frac{V E}{E W}$
   
   
   
   

   $\begin{array}{l} \quad\quad\text{If}\ AE\ \text{bisects}\ \angle WAV,\\\\ \quad\quad\dfrac{WE}{EV}=\dfrac{WA}{AV}\\\\ \Rightarrow\quad \dfrac{WE}{WA}=\dfrac{EV}{AV}\\\\ \Rightarrow\quad \dfrac{AV}{AW}=\dfrac{VE}{EW}\\\\ \therefore\quad \text{(a), (c), (d) are true and (b) is false.} \end{array}$

   
   


2. $AX$ bisects $\angle CAB$. Complete the following statements:
   
   (a) $A C: A B=\ldots$
   
   (b) $A B: A C=\ldots$
   
   (c) $X C: X B=\ldots$
   
   
   
   

   $\begin{array}{l} \text{(a)}\ \dfrac{AC}{AB}=\dfrac{CX}{XB}\\\\ \text{(b)}\ \dfrac{AB}{AC}=\dfrac{BX}{XC}\\\\ \text{(c)}\ \dfrac{XC}{XB}=\dfrac{AC}{AB} \end{array}$

   
   


3. $PT$ bisects $\angle RPS$. Complete the following statements:
   
   (a) $P Q: P R=\ldots$
   
   (b) $T R: P R=\ldots$
   
   (c) $Q R: T R=\ldots$
   
   
   
   

   $\begin{array}{l} \text{(a)}\ \dfrac{PQ}{PR}=\dfrac{TQ}{TR}\\\\ \text{(b)}\ \dfrac{TR}{PR}=\dfrac{TQ}{PQ}\\\\ \text{(b)}\ \dfrac{QR}{TR}=\dfrac{TQ-TR}{TR}=\dfrac{PQ-PR}{PR} \end{array}$

   
   


4. What can you say about the rays $AD$, $BE$ and $CF$?
   
   

   $\begin{array}{ll} \quad\quad AB = 6, AC = 8, BC = 10\\\\ \quad\quad\dfrac{AB}{AC}=\dfrac{6}{8}=\dfrac{3}{4}\\\\ \quad\quad\dfrac{AC}{BC}=\dfrac{8}{10}=\dfrac{4}{5}\\\\ \quad\quad\dfrac{BC}{AB}=\dfrac{10}{6}=\dfrac{5}{3}\\\\ \quad\quad\dfrac{BD}{DC}=\dfrac{\frac{30}{7}}{\frac{40}{7}}=\dfrac{3}{4}\\\\ \quad\quad\dfrac{AF}{BF}=\dfrac{\frac{8}{3}}{\frac{10}{3}}=\dfrac{4}{5}\\\\ \quad\quad\dfrac{CE}{AE}=\dfrac{5}{3}\\\\ \therefore\quad \dfrac{AB}{AC}=\dfrac{BD}{DC}\Rightarrow AD\ \text{bisects}\ \angle BAC\\\\ \therefore\quad \dfrac{AC}{BC}=\dfrac{AF}{BF}\Rightarrow CF\ \text{bisects}\ \angle ACB\\\\ \therefore\quad \dfrac{BC}{AB}=\dfrac{CE}{AE}\Rightarrow BE\ \text{bisects}\ \angle ABC \end{array}$

   
   


5. If $AD$ and $AE$ are bisectors of the interior and exterior angles at $A$ of $\triangle ABC$, then which of the following are true?
   
   (a) $\angle D A E=90^{\circ}$
   
   (b) $B D: D C=B C: C E$
   
   (c) $B D: D C=B E: C E$
   
   (d) $A D: A E=D C: C E$
   
   

   $\begin{array}{ll} \quad\quad (\bullet) + (\bullet) + (*) + (*) = 180^{\circ}\\\\ \quad\quad 2(\bullet) + 2(*) = 180^{\circ}\\\\ \quad\quad (\bullet) + (*) = 90^{\circ}\\\\ \therefore\quad \angle DAE = 90^{\circ}\\\\ \therefore\quad \text{(a) is true.}\\\\ \quad\quad AD\ \text {bisects}\ \angle BAC.\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}\\\\ \quad\quad AE\ \text {bisects}\ \angle CAF.\\\\ \therefore\quad \dfrac{BE}{CE}=\dfrac{AB}{AC}\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{BE}{CE}\\\\ \therefore\quad \text{(b) is false and (c) is true.}\\\\ \quad\quad \text {We cannot say that}\ (\bullet) = (*)\\\\ \therefore\quad \text{We cannot say that}\ AC\ \text {bisects}\ \angle DAE.\\\\ \therefore\quad \text{We cannot say that}\ \dfrac{AD}{AE}=\dfrac{DC}{CE}.\\\\ \therefore\quad \text{(d) is false.} \end{array}$

   
   


6. Find the value of 𝑥 in each of the following figures.
   
   (a)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{3}=\dfrac{10}{6}\\\\ \therefore\quad x=\dfrac{10}{6}\times 3=5 \end{array}$

   
   (b)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{9}=\dfrac{13}{8}\\\\ \therefore\quad x=\dfrac{13}{8}\times 9=\dfrac{117}{8}=14.625 \end{array}$

   
   (c)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{4-x}=\dfrac{8}{6}\\\\ \therefore\quad 3x=16-4x\\\\ \quad\quad 7x=16\\\\ \quad\quad x=\dfrac{16}{7}=2.286 \end{array}$

   
   (d)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{6-x}=\dfrac{4}{3}\\\\ \therefore\quad 3x=24-4x\\\\ \quad\quad 7x=24\\\\ \quad\quad x=\dfrac{24}{7}=3.429 \end{array}$

   
   (e)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{x-2}=\dfrac{7}{3}\\\\ \therefore\quad 3x=7x-14\\\\ \quad\quad 4x=14\\\\ \quad\quad x=\dfrac{7}{2}=3.5 \end{array}$

   
   (f)
   
   

   $\begin{array}{l} \quad\quad\dfrac{x}{x+2.3}=\dfrac{2.3}{3.8}\\\\ \therefore\quad \dfrac{10x}{10x+23}=\dfrac{23}{38}\\\\ \quad\quad 380x=230x+529\\\\ \quad\quad 150x=529\\\\ \quad\quad x=\dfrac{529}{150}=3.527 \end{array}$

   
   


7. Find the unknown marked lengths in the figure.
   
   

   $\begin{array}{l} \quad\quad CD\ \text{bisects}\ \angle ACB\\\\ \therefore\quad \dfrac{AD}{DB}=\dfrac{AC}{BC}\\\\ \therefore\quad \dfrac{x}{24}=\dfrac{20}{30}\\\\ \therefore\quad x=16\\\\ \quad\quad AE\ \text{bisects}\ \angle CAF\\\\ \therefore\quad \dfrac{CE}{BE}=\dfrac{AC}{AB}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{20}{x+24}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{20}{40}\\\\ \quad\quad \dfrac{y}{30+y}=\dfrac{1}{2}\\\\ \quad\quad 2y=30+y\\\\ \therefore\quad y=30\\\\ \end{array}$

   
   


8. $A B=12$ cm, $B C=9$ cm $C A=7$ cm. $B D$ bisects $\angle B$ and $A G=A D$, $C H=C D .$ Calculate $B G, B H$. Does $G H \parallel A C$ ?
   
   

   $\begin{array} \quad\quad \text{Let}\ AG=AD=y\ \text{and}\ CH=CD=x.\\\\ \therefore\quad x+y=7\Rightarrow y=7-x\\\\ \quad\quad \text{Since}\ BD\ \text{bisects}\ \angle ABC\\\\ \therefore\quad \dfrac{AD}{DC}=\dfrac{AB}{BC}\\\\ \quad\quad \dfrac{y}{x}=\dfrac{12}{9}\\\\ \quad\quad \dfrac{7-x}{x}=\dfrac{4}{3}\\\\ \therefore\quad 21-3x = 4x\\\\ \quad\quad 7x=21\\\\ \therefore\quad x = 3\\\\ \therefore\quad y = 7-x=7-3=4 \end{array}$

   
   


9. In $\triangle A B C, D E \parallel B C$, $A D=2.7$ cm, $D B=1.8$ cm and $B C=3$ cm. Prove that $B E$ bisects $\angle A B C$.
   
   

   $\begin{array} \quad\quad DE\parallel BC\\\\ \therefore\quad \dfrac{AE}{EC}=\dfrac{AD}{DB}\\\\ \quad\quad \dfrac{AE}{EC}=\dfrac{2.7}{1.8}=\dfrac{3}{2}\\\\ \quad\quad \text{But}\ \dfrac{AB}{BC}=\dfrac{2.7+1.8}{3}=\dfrac{3}{2}\\\\ \therefore\quad \dfrac{AE}{EC}=\dfrac{AB}{BC}\\\\ \therefore\quad BE\ \text{bisects}\ \angle ABC. \end{array}$

   
   


10. In a parallelogram $A B C D$, $A B=3.6$ cm, $B C=2.7$ cm, $A X=3.2$ cm, $X C=2.4$ cm. Prove that $\triangle B C Y$ is isosceles.
    
    

    $\begin{array}{l} \quad\quad \dfrac{AX}{XC}=\dfrac{3.2}{2.4}=\dfrac{4}{3}\\\\ \quad\quad \dfrac{AB}{BC}=\dfrac{3.6}{2.7}=\dfrac{4}{3}\\\\ \therefore\quad \dfrac{AX}{XC}=\dfrac{AB}{BC}\\\\ \therefore\quad BY\ \text{bisects}\ \angle ABC.\\\\ \therefore\quad \alpha = \beta\\\\ \quad\quad \text{Since}\ DC\parallel AB,\\\\ \quad\quad \alpha = \gamma\quad (\text{alternate}\ \angle\text{s})\\\\ \therefore\quad \beta =\gamma\\\\ \therefore\quad \triangle BCY \text{is isosceles.} \end{array}$ Angle bisector theorem အောက်တွင် မေးထားသော မေးခွန်း ဖြစ်သောကြောင့် Angle bisector theorem ဖြင့် သက်သေပြခြင်း ဖြစ်သော်လည်း သဏ္ဌာန်တူ ဥပဒေသကို သုံး၍လည်း သက်သေပြနိုင်ပါသည်။ $\begin{array}{l} \quad\quad \text{Since}\ DC\parallel AB,\\\\ \quad\quad \triangle AXB \sim \triangle CXY\\\\ \therefore\quad \dfrac{AB}{CY}=\dfrac{AX}{CX}\\\\ \therefore\quad \dfrac{3.6}{CY}=\dfrac{3.2}{2.4}\\\\ \quad\quad CY=\dfrac{3}{4} \times 3.6 = 2.7\ \text{cm}\\\\ \therefore\quad BC = CY\\\\ \therefore\quad \triangle BCY \text{is isosceles.} \end{array}$

    
    


11. Calculate $𝐵𝐷$ and $𝐷𝐶$ in terms of $𝑎, 𝑏, 𝑐$.
    
    

    $\begin{array} \quad\quad BD + DC = a\Rightarrow DC = a -BD\\\\ \quad\quad BY\ \text{bisects}\ \angle ABC.\\\\ \therefore\quad \dfrac{BD}{DC}=\dfrac{AB}{AC}\\\\ \therefore\quad \dfrac{BD}{a - BD}=\dfrac{c}{b}\\\\ \quad\quad b\cdot BD=ac-c\cdot BD\\\\ \quad\quad b\cdot BD + c\cdot BD=ac\\\\ \quad\quad (b+c) BD =ac\\\\ \therefore\quad BD=\dfrac{ac}{b+c}\\\\ \therefore\quad DC=a-\dfrac{ac}{b+c}=\dfrac{ab}{b+c} \end{array}$

    
    


12. Given : $A H$ bisects $\angle B A C$ in $\triangle A B C$. $E H \parallel A C$
    
    Prove : $\displaystyle \frac{B E}{E A}=\displaystyle \frac{B A}{A C}$
    
    

    $\begin{array} \quad\quad \text{Since}\ AH\ \text{bisects}\ \angle ABC,\\\\ \quad\quad \dfrac{BA}{AC}=\dfrac{BH}{HC}\\\\ \quad\quad \text{Since}\ EH\parallel AC,\\\\ \quad\quad \dfrac{BE}{EA}=\dfrac{BH}{HC}\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BA}{AC} \end{array}$

    
    


13. Given : In $\triangle A B C, B M=M C$.
    
                $M X$ bisects $\angle A M B$ .
    
                $M Y $ bisects $\angle A M C$.
    
    Prove : $X Y \parallel B C$ .
    
    

    $\begin{array} \quad\quad \text{Let}\ AM=x\ \text{and}\ BM=MC=y,\\\\ \quad\quad \text{Since}\ MX\ \text{bisects}\ \angle AMB,\\\\ \quad\quad \dfrac{AX}{XB}=\dfrac{AM}{BM}=\dfrac{x}{y}\\\\ \quad\quad \text{Similarly}\ MY\ \text{bisects}\ \angle AMC,\\\\ \quad\quad \dfrac{AY}{YC}=\dfrac{AM}{CM}=\dfrac{x}{y}\\\\ \therefore\quad \dfrac{AX}{XB}=\dfrac{AY}{YC}\\\\ \therefore\quad XY\parallel BC \end{array}$

    
    


14. Given : In $\triangle A B C$, $\angle A=2 \angle C$,
    
                $A D$ bisects $\angle B A C$ and
    
                $D E$ bisects $\angle A D B$.
    
    Prove $: \displaystyle \frac{B E}{E A}=\displaystyle \frac{B A}{A C}$
    
    

    $\begin{array}{l} \quad\quad \text{Since}\ \angle A=2\angle C,\\\\ \quad\quad \angle C=\dfrac{1}{2}\angle A\\\\ \therefore\quad \angle C=\angle DAC\\\\ \therefore\quad \triangle DCA\ \text{is isosceles with}\ DC = DA\\\\ \quad\quad \text{Since}\ AD\ \text{bisects}\ \angle BAC,\\\\ \quad\quad \dfrac{BD}{DC}=\dfrac{BA}{AC}\Rightarrow \dfrac{BD}{DA}=\dfrac{BA}{AC}\quad (\because DC=DA)\\\\ \quad\quad \text{Similarly}\ DE\ \text{bisects}\ \angle ADB,\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BD}{DA}\\\\ \therefore\quad \dfrac{BE}{EA}=\dfrac{BA}{AC} \end{array}$

    
    

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